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Why in the code "456"+1, output is "56" [duplicate]
(3 answers)
What is the answer when integer added to string constant in C language?
(4 answers)
Closed last year.
cout<<"ccccc"+2;
Output:
ccc
I tried searching for it online and I know it is a very dumb question but couldn't find anything anywhere. Please if someone could help me out.
"ccccc"+2;
"ccccc" decays to the const char * pointer referencing the first character of the string literal "ccccc". When you add 2 to it, the result references the third element of the string literal.
It is the same as:
const char *cptr = "ccccc";
cptr += 2;
cout << cptr;
When you wrote:
cout<<"ccccc"+2;
The following things happen(to note here):
"ccccc" is a string literal. In particular, it is of type const char[6].
Now, this string literal decays to a pointer to const char which is nothing but const char* due to type decay. Note that the decayed const char* that we have now is pointing to the first character of the string literal.
Next, 2 is added to that decayed pointer's value. This means that now, after adding 2, the const char* is pointing to the third character of the string literal.
The suitable overloaded operator<< is called using this const char*. And since this const char* is pointing to the third character of the string literal, you get the output you observe.
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What is the difference between char s[] and char *s?
(14 answers)
Closed 1 year ago.
char* a = "string"; /*"string" is a string literal, thus modifying
value isn't allowed, but then why */
char b[] = "string1"; /*"string1" is
also a string literal but why modification of this is allowed? */
a[1] = 's'; //this is not allowed b[1] = 'p'; //this is allowed
Why can char array be modified when it is clearly pointing to a string literal?
when it is clearly pointing to a string literal?
No. Given char b[] = "string1";, b is not a pointer pointing to the string literal, but a char array containing copy of "string1". The modification on the string literal leads to UB, but modification on the char array is fine.
String literals can be used to initialize character arrays. If an array is initialized like char str[] = "foo";, str will contain a copy of the string "foo".
BTW char* a = "string"; is not allowed since C++11; you have to write it as const char* a = "string";.
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Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"?
(19 answers)
Closed 2 years ago.
#include <stdio.h>
#include <string.h>
int main()
{
char t1string[] = "test";
printf("First String value is %s\n", t1string);
strcpy(t1string, "lgs");
printf("Modified First String value is %s\n", t1string);
char * t2string = "test";
printf("Second String value is %s\n", t2string);
strcpy(t2string, "lgs");
printf("Modified Second String value is %s\n", t2string);
return 0;
}
Looking at the above pair of strcpy usage, I am trying to understand why the strcpy used in the second instance does not work, or is it something specifically wrong (with either my beginner level understanding or code blocks)? Or is it just telling me I am stupid for not simply doing t2string = "lgs";?
Whenever you see a string literal such as "test" in C, that is a const char[5] array which decays for most all purposes into const char *, aka a pointer to a constant character array. The letters 't', 'e', 's', 't' cannot be changed. They are stored in the same area of the program as the program instructions. Those cannot be changed either.
When you declare an array like char s[] = "test"; the constant characters "test" are copied into the array s, exactly as if you had called memcpy(s, "test", 5); From their new location in s they can be modified.
As a backward compatibility support C allows creating char * pointers to const char * arrays. But your program will still crash if you attempt to write into that constant array.
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how is char * to string literal valid?
(5 answers)
Closed 5 years ago.
Using visual studio, I declared a pointer of type char * and assigned to it a string literal. I then hovered the mouse over the string literal and it displayed its type: (const char [4])"abc".
How is this allowed? it compiles without warnings or errors, whilst assigning to the pointer an array of type const char [] fails, for obvious reasons, with an error message:
a value of type "const char *" cannot be assigned to an entity of type "char *"
So, why is it allowed for string literals?
int main(void)
{
char *p = "abc"; // no error here
const char str[] = "abc";
//p = str; This line generates an error
return 0;
}
EDIT: answer updated to incorporate info from Story Teller's comment.
In the olden days, const didn't exist, and people wrote things like char* p = "abc" all the time. As long as the didn't then do something like p[0] = 'z', their program worked. To remain compatible with such code, some compilers allow string literals to be assigned to non-const pointers if you don't ask the compiler to be super-strict. If you take advantage of this feature, you still should not ACTUALLY modify the string.
im realy confused about const char * and char *.
I know in char * when we want to modify the content, we need to do something like this
const char * temp = "Hello world";
char * str = new char[strlen(temp) + 1];
memcpy(str, temp, strlen(temp));
str[strlen(temp) + 1] = '\0';
and if we want to use something like this
char * str = "xxx";
char * str2 = "xts";
str = str2;
we get compiler warning. it's ok I know when i want to change char * I have to use something memory copy. but about const char * im realy confused. in const char * I can use this
const char * str = "Hello";
const char * str2 = "World";
str = str2; // and now str is Hello
and I have no compiler error ! why ? why we use memory copy when is not const and in const we only use equal operator ! and done !... how possible? is it ok to just use equal in const? no problem happen later?
As other answers say, you should distinguish pointers and bytes they point to.
Both types of pointers, char * and const char *, can be changed, that is, "redirected" to point to different bytes. However, if you want to change the bytes (characters) of the strings, you cannot use const char *.
So, if you have string literals "Hello" and "World" in your program, you can assign them to pointers, and printing the pointer will print the corresponding literal. However, to do anything non-trivial (e.g. change Hello to HELLO), you will need non-const pointers.
Another example: with some pointer manipulation, you can remove leading bytes from a string literal:
const char* str = "Hello";
std::cout << str; // Hello
str = str + 2;
std::cout << str; // llo
However, if you want to extract a substring, or do any other transformation on a string, you should reallocate it, and for that you need a non-const pointer.
BTW since you are using C++, you can use std::string, which makes it easier to work with strings. It reallocates strings without your intervention:
#include <string>
std::string str("Hello");
str = str.substr(1, 3);
std::cout << str; // ell
This is a confusing hangover from the days of early C. Early C didn't have const, so string literals were "char *". They remained char * to avoid breaking old code, but they became non-modifiable, so const char * in all but name. So modern C++ either warns or gives an error (to be strictly conforming) when the const is omitted.
Your memcpy missed the trailing nul byte, incidentally. Use strcpy() to copy a string, that's the right function with the right name. You can create a string in read/write memory by use of the
char rwstring[] = "I am writeable";
syntax.
That is cause your variables are just a pointers *. You're not modifiying their contents, but where they are pointing to.
char * a = "asd";
char * b = "qwe";
a = b;
now you threw away the contents of a. Now a and b points to the same place. If you modify one, both are modified.
In other words. Pointers are never constants (mostly). your const predicate in a pointer variable does not means nothing to the pointer.
The real difference is that the pointer (that is not const) is pointing to a const variable. and when you change the pointer it will be point to ANOTHER NEW const variable. That is why const has no effect on simple pointers.
Note: You can achieve different behaviours with pointers and const with more complex scenario. But with simple as it, it mostly has no effect.
Citing Malcolm McLean:
This is a confusing hangover from the days of early C. Early C didn't have const, so string literals were "char *". They remained char * to avoid breaking old code, but they became non-modifiable, so const char * in all but name.
Actually, string literals are not pointers, but arrays, this is why sizeof("hello world") works as a charm (yields 12, the terminating null character is included, in contrast to strlen...). Apart from this small detail, above statement is correct for good old C even in these days.
In C++, though, string literals have been arrays of constant characters (char const[]) right from the start:
C++ standard, 5.13.5.8:
Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration.
(Emphasised by me.) In general, you are not allowed to assign pointer to const to pointer to non-const:
char const* s = "hello";
char ss = s;
This will fail to compile. Assigning string literals to pointer to non-const should normally fail, too, as the standard explicitly states in C.1.1, subclause 5.13.5:
Change: String literals made const.
The type of a string literal is changed from “array of char” to “array of const char”.
[...]char* p = "abc"; // valid in C, invalid in C++
Still, string literal assignement to pointer to non-const is commonly accepted by compilers (as an extension!), probably to retain compatibility to C. As this is, according to the standard, invalid, the compiler yields a warning, at least...